# Lorentz Covariant formula for Noether Charges in QFT

+ 2 like - 0 dislike
352 views

I'm looking for a Lorentz covariant expression of Noether charges and I found this article: https://arxiv.org/abs/hep-th/0701268, section II-A in particular.

Consider specifically eq. (20-21), they claim: $$Q_\mu=\frac{1}{2}(\phi, P_\mu \rhd \phi),$$ $Q$ is the conserved charge,  "$\rhd$" is just an "acting on" symbol and the inner product is defined by ($\varepsilon$ is the sign function): $$(\phi_1, \phi_2)=\int d^4p \, \delta(p^2-m^2) \varepsilon(p_0)\tilde{\phi}_1^*(-p)\tilde{\phi}_2(p).$$
$\tilde{\phi}$ is the fourier transform of $\phi$.

Hence the Noether Charge is \label{1} \tag{1}
Q_\mu =
\frac{1}{2}\int d^4p \, \delta(p^2-m^2) \varepsilon(p_0)p_\mu\tilde{\phi}^*(-p)\tilde{\phi}(p).

Now I'm struggling to get the well known quantised expression in QFT: $$Q_\mu= \int d^3p\,\, p_\mu \,\,a^\dagger (\vec{p}) \,a (\vec{p}), \,\,\,\,[a^\dagger (\vec{p}), \,a (\vec{q})]=-\delta^3 (\vec{p}-\vec{q})$$
from (1), by plugging in usual scalar klein gordon field with creation and annihilation operators..

If I'm not wrong (1) in coordinate space looks like $$\int d^4x \,\,d^4y \,\, \phi (x) \Delta (x-y)\left( -i\frac{\partial \phi(y)}{\partial y^\mu}\right)=Q_\mu, \tag{2}$$
where $\Delta$ is the usual commutator function $$\Delta (x-y)=\int d^4p \,\, \varepsilon(p_0) \delta (p^2-m^2)e^{-ip\cdot (x-y)}.$$ Just by substituting in (2) $$\phi(x)= \int \frac{d^3p}{\sqrt{2\omega}_\vec{p}}(a(\vec{p}) e^{-ip\cdot x}+a^\dagger(\vec{p}) e^{ip\cdot x} )$$ I don't seem to be getting the right answer.

Maybe I'm doing some calculation wrong or misinterpreted the article. Any help would be greatly appreciated!

My try:

For instance writing $\phi(x)=\int d^4p \,\, a(p) \delta(p^2-m^2) e^{-ip \cdot x}$ then $\tilde{\phi}(p)=a(p)\delta(p^2-m^2)$ and (1) becomes $$Q_\mu= \frac{1}{2} \int d^4p \,\, \varepsilon(p_0) \, p_\mu a(-p)a(p)\,\,\delta(p^2-m^2)\delta(p^2-m^2)\delta(p^2-m^2).$$
Is this right? How to work out the three deltas? I could use the identity $\delta(x)f(x)=\delta(x)f(0)$ with  $f=\delta$ twice to get$$\delta(p^2-m^2)\delta(p^2-m^2)\delta(p^2-m^2)=\delta(p^2-m^2)\delta(0)\delta(0)=\delta(p^2-m^2)\cdot \mathcal{S},$$ where $\mathcal{S}$ is an (infinite) surface contribution which I currenlty fail to see how cancels out.. what am I missing? Maybe they're using a different convention of their fourier transforms?

 Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead. To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL. Please consult the FAQ for as to how to format your post. This is the answer box; if you want to write a comment instead, please use the 'add comment' button. Live preview (may slow down editor)   Preview Your name to display (optional): Email me at this address if my answer is selected or commented on: Privacy: Your email address will only be used for sending these notifications. Anti-spam verification: If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:p$\hbar$ysicsOverf$\varnothing$owThen drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds). To avoid this verification in future, please log in or register.